Chapter 3 C 2 H 4 O2. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives. C x H y Oz

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Announcements HOUR EXAM 1 --Want me to do recitation again? July 18 6-7:30PM --Skip Combustion Analysis & Isomers (p.82-83 in Principles of Chemistry Text) See me if you donʼt understand!

1 Announcements HOUR EXAM 1 --Want me to do recitation again? July :30PM --Skip Combustion Analysis & Isomers (p in Principles of Chemistry Text) See me if you donʼt understand! Chapter 3 Relationships, Stoichiometry and Chemical s Chapter 3: 6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32, 33, 38, 40, 41, 43, 45, 49, 51, 65, 66, 69, 71, 73, 75, 83, 85, 89, 93, 95, (Principles of Chemistry) Chapter 3: 3.7, 3.11, 3.13, 3.15, 3.17, 3.19, 3.26, 3.29, 3.34,3.36, 3.38, 3.42, 3.45, 3.52, 3.56, 3.62, 3.64, 3.66, 3.72, 3.74, 3.82, 3.88, 3.93, 3.95, 3.97, 3.99, 3.101,3.114, 3.118, (4th Chemistry Nature of Matter) Learning Objectives 1.Understand relative atomic masses, average isotopic mass. 2.Connect the dots between amu & grams & the periodic table via the mole. 3.Compute a molecular & molar mass of a substance from a formula. 4.Using factor label method to convert between grams<=>moles<=>molecules 5.% To Empirical and Vis-versa 6. Balancing equations and mastering stoichiometry 7. Limiting Reagent, Yields, Solution Stoichiometry A chemical formula determines the % mass of each element in a compound. n x molar mass of element molar mass of compound x 100% n is the number of moles of an element in 1 mole of the compound. C 2 H 4 O2 2 x (12.01 g) %C = x 100% = 40.0% g 4 x (1.008 g) %H = x 100% = 6.714% g 2 x (16.00 g) %O = x 100% = 53.28% g 40.0% % % = 100.0% A laboratory technique called elemental analysis can determine the % by mass of each element in a compound. We can compute the empirical formula of any compound with that information. % Element %C = a% %H = b% %O = c% Molar Empirical C x H y Oz Empirical and s The formula of a compound as it actually exists according to experimental data. It is a multiple of the empirical formula. Molar CH 2O C 2H 4O 2 C 3H 6O 3 C 4H 8O 4 C 5H 10O Empirical The simplest formula for a compound that gives rise to the smallest set of whole numbers of atoms. CH 2O

2 Compounds that have the same % mass of its elements have the same empirical formula! Name formaldehyde acetic acid lactic acid erythrose ribose glucose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6 Whole-Number Multiple M (g/mol) All have the same % by mass! Use or Function disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle part of sugar metabolism component of nucleic acids and B 2 major energy source of the cell 40.0% C 6.71% H 53.3% O. Determining the Empirical from es of Elements (Elemental Analysis) Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. Determine the empirical formula and the name of this compound? SOLUTION: 2.82 g Na x mol Na = mol Na g Na 4.35 g Cl x mol Cl = mol Cl g Cl 7.83 g O x mol O g O = mol O Na 1 Cl 1 O 3.98 NaClO 4 NaClO 4 is sodium perchlorate. Determining the Empirical From % Data Problem Solving Method Convert % to grams (Assume g) Convert each element in grams to moles Divide Moles by the smallest number of moles Simplify the Mole Ratio May be necessary to multiply (or divide) all moles by a common term to convert to the entire formula to whole numbers. Use Molar to Determine (molar mass must be given). Determining a from Elemental Analysis and Molar Dibutyl succinate is an insect repellent used against household ants and roaches. Elemental analysis or analysis indicates that % mass of the composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? formula Step 1: Determine the mass of each element. Assume a 100 gram sample and use the given data we have: C g H 9.63 g O g Step 2: Convert masses to amounts in moles. Step 3: Write a tentative empirical formula. C 5.21 H 9.55 O 1.74 Step 4: Convert to small whole numbers. Divide by smallest number of moles C 2.99 H 5.49 O Step 5: Convert to a small whole number ratio. Multiply by 2 to get C 5.98 H O 2 The empirical formula is C 6 H 11 O 2 Empirical formula mass = 6(12.01) (11) + 2 (16.00) = 115 u. Step 6: Now using the empirical formula mass and molecular mass together determine the molecular formula. Empirical formula mass is 115 u. formula mass is 230 u. n = mass/empirical mass = 230 amu/115 amu = 2 The molecular formula is C 12 H 22 O 4

Determining a from Elemental Analysis and Molar During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue.

3 Determining a from Elemental Analysis and Molar During physical activity, lactic acid (M = g/mol) forms in muscle tissue and is responsible for muscle soreness at fatigue. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. Understand what is asked: What is the formula CxHyOz Percent Empirical Determining a from Elemental Analysis and Molar 1. Assume there are 100. g of lactic acid then use % mass: 40.0 g C 1 mol C 6.71 g H 1 mol H 53.3 g O 12.01g C g H 1 mol O g O 3.33 mol C 6.66 mol H 3.33 mol O 2. The red numbers are the number of moles of atoms in lactic acid. This is what we use in the formula C 3.33 H 6.66 O 3.33 C 3.33 H 6.66 O molar mass of lactate mass of CH 2 O CH 2 O g g empirical formula 3. The molecular formula will be a whole number multiple of the empirical formula determined BY THE MOLAR MASS ratio 3 C 3H 6 O 3 is the molecular formula Chemical equations are symbolic representations of what happens in a chemical reaction. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) All stoichiometry calculations are based on information contained in a balanced chemical equation. + Reactants Yields Products SMALL 1 formula unit + 3 molecules => 2 atoms + 3 molecules amu amu => amu amu Balanced equations = Conservation of! Balanced equations = Correct Stoichiometry! Phases specified: (s) for solids, (l) for liquids, (g) for gases, (aq) for aqueous. BIG 1 mol Fe2O3 + 3 mol CO => 2 mol Fe + 3 mol CO g g => g g Only balanced chemical equations contain useful stoichiometric conversion factors. We have to know how to balance equations! It s trial and error! NH3 + O2 ===> NO + H2 1 moles NH3 = 1 moles O2 Not Balanced Na3PO4(aq) + HCl(aq) ==> H3PO4(aq) + NaCl(aq) 6NH3 + 3O2 ===> 6NO + 9H2 Balance first! Ba(OH)2(aq) + HCl(aq) ==> H2O(l) + BaCl2(aq) Correct Stoichiometric Conversion Factors 6 mol NH3 = 3 mol O2 6 mol NH3 = 6 mol NO 6 mol NH3 = 9 mol H2 3 mol O2 = 6 mol NO 3 mol O2 = 9 mol H2 6 mol NO = 9 mol H2 C7H14 + O2 ===> H2O(l) + CO2(g)

4 Solutions Na3PO4(aq) + 3HCl(aq) ====> H3PO4(aq) + 3NaCl(aq) Ba(OH)2(aq) + 2HCl(aq) ====> 2H2O(l) + BaCl2(aq) 2C7H O2 ====> 14H2O(l) + 14CO2(g) Some are tougher than others...practice! Example: Ethane, C2H6, reacts (is combusted are key words) with O2 to form CO2 and H2O. Write a balanced equation for this reaction. 1. Write the correct formula(s) for reactants and products. CO 2 + H 2 O 2. Start by balancing those elements that appear in only one reactant and one product. CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 6 hydrogen on left 2CO 2 + H 2 O 2 hydrogen on right 2CO 2 + 3H 2 O multiply H 2 O by 3 4. Balance those elements that appear in two or more reactants or products. 2 O on left 4 O 2CO 2 + 3H 2 O + 3 O = 7 oxygen on right 4. Balance those elements that appear in two or more reactants or products. 2CO 2 + 3H 2 O 2 O on left 4 O C 2 H O 2 2CO 2 + 3H 2 O 2 2C 2 H 6 + 7O O = 7 oxygen on right 4CO 2 + 6H 2 O multiply O 2 by 7 2 remove fraction multiply by 2 C 2 H O 2 2CO 2 + 3H 2 O 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O multiply O 2 by 7 2 remove fraction multiply both sides by 2 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O To learn how to balance equations quickly for Exams you have to practice! Try it. A balanced chemical equation contains all the information we need to do many calculations! Na3PO4(aq) + HCl(aq) ===> H3PO4(aq) + NaCl(aq) Ba(OH)2(aq) + HCl(aq) ===> H2O(l) + BaCl2(aq) C7H14 + O2 ====> H2O(l) + CO2(g) Na2SO4 (s) + C(s) Na2S (s) + CO(g) Na + H2O NaOH + H2 (g) There is a lot of information here! 1 mol Fe2O3 => 3 mol CO 1 mol Fe2O3 => 3 mol 1 mol Fe2O3 => 2 mol Fe CO2 3 mol CO => 2 mol Fe 3 mol CO => 3 mol CO2 Mg3N2(s) + H2O(l) H2S (g) + SO2(g) Mg(OH)2 (s) + NH3(g) S (s) + H2O(l) not balanced = NOT RIGHT! Fe2O3 (s) + CO(g) Fe(s) + CO2(g)

5 There is lots of information in a balanced chemical equation! molecules mass of atoms amount (mol) mass (g) C 3 H 8 (g) + 5O 2 (g) 1 molecule C 3 H molecules O amu C 3 H amu O 2 1 mol C 3 H mol O g C 3 H g O 2 3CO 2 (g) + 4H 2 O(g) 3 molecules CO molecules H 2 O amu CO amu H 2 O 3 mol CO mol H 2 O g CO g H 2 O Steps to mastering stoichiometry! 1. Always write a balanced chemical equation. 2. Work in moles---not masses...we need to count? 3. Use dimensional analysis correctly. Grams of Reactant Molar Moles of Reactant balanced equation Grams of Product Moles of Product Molar total mass (g) g g Iron III oxide reacts with carbon monoxide as shown below. How many CO molecules are required to react with 25 formula units of Fe 2 O 3 as shown below in the balanced equation? How many CO molecules are required to react with 25 formula units of Fe 2 O 3 as shown below in the balanced equation? Fe2O3 (s) + CO(g) Fe(s) + CO2(g) Translate words to formula THE MAGIC 1. Balance chemical equation first! 2. What does the question want? 3. Find stoichiometric factors 4. Use the factor-label method and solve 5. Be mindful of significant figures 6. Check answer Balance First!! How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide? What mass of CO is required to react with 146 g of iron (III) oxide? # Fe atoms = = fu Fe 2 O 3 2 F e atoms 1 fu Fe 2 O 3 = F e atoms

What mass (in grams) of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

It tells you one reactant is in excess and the other is not! What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

We need to count the number of reactants to figure out what will limit the amount of cars produced--it s not how much they weigh!

6 What mass (in grams) of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? What mass of carbon dioxide can be produced by the reaction of mole of iron (III) oxide with excess carbon monoxide? will run out first excess key riff! It tells you one reactant is in excess and the other is not! What mass of carbon dioxide can be produced by the reaction of mole of iron (III) oxide with excess carbon monoxide? The limiting reagent is the reactant that is runs out (is consumed) and determines the quantity or amount of product that can be formed. We need to count the number of reactants to figure out what will limit the amount of cars produced--it s not how much they weigh! We deal with limiting reagents all the time, we just don t give it a buzzword like chemists do. Reactants Product Do You Understand Limiting Reagent II? If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? Notice the absence of excess and there are two reactant masses in the problem = limiting reagent! 168/2 = /4 = 82 There are two ways you can calculate the answer. Method 1: For both reactants, use balanced equation & implied stoichiometric factors and compute the amount of any product formed (I prefer and teach this method!) Method 2: Pick one of the reactant and compute how much of the other reactant you need and compare with the given amount.

7 If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? CH 4 (g) + 2 O 2 (g)!!" CO 2 (g) + 2 H 2 O(g) Method 1 (Calculating the # moles of product formed by each reactant to determine which reactant makes the least amount) 1. Let s use the amount of CO2 formed as our yardstick of how much product can be made (we could chose H2O). mol CO 2 = 25.0 g CH 4 1 mol CH g CH 4 1 mol CO 2 1 mol CH 4 = mol CO 2 mol CO 2 = 40.0 g O 2 1 mol O g O 2 1 mol CO 2 2 mol O 2 = mol CO 2 O2 must be the limiting reagent as the amount of CO2 produced is the least amount of product! If 25.0 g CH 4 is combusted with 40.0 g O 2, which reactant is the limiting reactant? CH 4 (g) + 2 O 2 (g)!!" CO 2 (g) + 2 H 2 O(g) Method 1I (Directly comparing amounts of reactants given in the problem to which is the limiting reagent--less steps) g O 2 needed = 25.0 g CH 4 1 mol CH g CH 4 2 mol O 2 1 mol CH g O 2 1 mol O 2 = g O 2 O2 is the limiting reagent as we need g of it but we are are only given 40.0 g O2! Thus, the amount of product that can be formed is determined by the amount of O2 not by the amount of methane, CH4.

Chapter 3. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives MOLE

Chapter 3. Mass Relationships, Stoichiometry and Chemical Formulas. Announcements. Learning Objectives MOLE Announcements HOUR EXAM 1 --Critical To Understand Chapter 3 July 18 6-7:30PM --Skip Combustion Analysis & Isomers (p.82-83 in Principles of Chemistry Text) See me if you donʼt understand! Chapter 3 Relationships,